我开始使用python,我尝试使用一个二维列表,我开始在每个地方都用相同的变量填充。我想到了这个:

def initialize_twodlist(foo):
    twod_list = []
    new = []
    for i in range (0, 10):
        for j in range (0, 10):
            new.append(foo)
        twod_list.append(new)
        new = []

它给出了预期的结果,但感觉像是一种变通方法。有更简单/更短/更优雅的方法吗?


当前回答

错误方式:[[None*m]*n]

>>> m, n = map(int, raw_input().split())
5 5
>>> x[0][0] = 34
>>> x
[[34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None]]
>>> id(x[0][0])
140416461589776
>>> id(x[3][0])
140416461589776

使用这种方法,python不允许为外层列创建不同的地址空间,这将导致各种超出您预期的错误行为。

正确方法,但有例外:

y = [[0 for i in range(m)] for j in range(n)]
>>> id(y[0][0]) == id(y[1][0])
False

这是一个很好的方法,但如果您将默认值设置为None,则会有例外

>>> r = [[None for i in range(5)] for j in range(5)]
>>> r
[[None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None]]
>>> id(r[0][0]) == id(r[2][0])
True

因此,使用这种方法正确地设置默认值。

绝对正确的:

跟着麦克风的双循环回复。

其他回答

错误方式:[[None*m]*n]

>>> m, n = map(int, raw_input().split())
5 5
>>> x[0][0] = 34
>>> x
[[34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None], [34, None, None, None, None]]
>>> id(x[0][0])
140416461589776
>>> id(x[3][0])
140416461589776

使用这种方法,python不允许为外层列创建不同的地址空间,这将导致各种超出您预期的错误行为。

正确方法,但有例外:

y = [[0 for i in range(m)] for j in range(n)]
>>> id(y[0][0]) == id(y[1][0])
False

这是一个很好的方法,但如果您将默认值设置为None,则会有例外

>>> r = [[None for i in range(5)] for j in range(5)]
>>> r
[[None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None], [None, None, None, None, None]]
>>> id(r[0][0]) == id(r[2][0])
True

因此,使用这种方法正确地设置默认值。

绝对正确的:

跟着麦克风的双循环回复。

from random import randint
l = []

for i in range(10):
    k=[]
    for j in range(10):
        a= randint(1,100)
        k.append(a)

    l.append(k)




print(l)
print(max(l[2]))

b = []
for i in range(10):
    a = l[i][5]
    b.append(a)

print(min(b))

你可以使用列表推导式:

x = [[foo for i in range(10)] for j in range(10)]
# x is now a 10x10 array of 'foo' (which can depend on i and j if you want)
twod_list = [[foo for _ in range(m)] for _ in range(n)]

n是行数,m是列数,foo是值。

通常,当你想要多维数组时,你不想要一个列表的列表,而是一个numpy数组,或者可能是一个dict。

例如,使用numpy可以执行如下操作

import numpy
a = numpy.empty((10, 10))
a.fill(foo)